Tính $\int_2^4\left(x+\dfrac{1}{x}\right)^2 \mathrm{~d} x$.

Ta có $\int_2^4\left(x+\dfrac{1}{x}\right)^2 \mathrm{~d} x=\int_2^4\left(x^2+2+\dfrac{1}{x^2}\right) \mathrm{d} x=\left.\left(\dfrac{x^3}{3}+2 x-\dfrac{1}{x}\right)\right|_2 ^4$ $=\left(\dfrac{4^3}{3}+8-\dfrac{1}{4}\right)-\left(\dfrac{2^3}{3}+4-\dfrac{1}{2}\right)=\dfrac{275}{12}$.